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otez555 [7]
3 years ago
15

4. A 62.0-kg person, standing on the diving board, dives straight down into the water. Just before striking the water, her speed

is 5.50 m/s. At a time of 1.65 s after she enters the water, her speed is reduced to 1.10 m/s. What is the net average force (magnitude and direction) that acts on her when she is in the water?
Physics
1 answer:
Alekssandra [29.7K]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the Moment. The moment in terms of the Force and the time can be expressed as

\Delta P = F\Delta t

F = Force

\Delta t = Time

At the same time the moment can be expressed in terms of mass and velocity, mathematically it can be given as

P = m \Delta v

Where

m = Mass

\Delta v = Change in velocity

Our values are given as

\Delta t=1.65s

By equating the two equations we can find the Force,

F\Delta t = m\Delta v

F = \frac{m\Delta v}{\Delta t}

F = \frac{62(1.1-5.5)}{1.65}

Therefore, the net average force will be:

F = - 165N

The negative symbol indicates that the direction of the force is upwards.

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The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

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Pressure is the amount of force per unit area. In formula it is,

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